Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 44

Answer

$y'=\dfrac{mx\cos mx-\sin mx}{x^{2}}$

Work Step by Step

$y=\dfrac{\sin mx}{x}$ Start the differentiation process by using the quotient rule: $y'=\dfrac{(x)(\sin mx)'-(x)'(\sin mx)}{x^{2}}=...$ Apply the chain rule to evaluate the indicated derivatives and simplify: $...=\dfrac{(x)(\cos mx)(mx)'-\sin mx}{x^{2}}=...$ $...=\dfrac{mx\cos mx-\sin mx}{x^{2}}$
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